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beks73 [17]
4 years ago
13

Electromagnetic waves differ only in their _____, their energy, and their frequency.

Physics
1 answer:
Tanya [424]4 years ago
6 0

Answer:

Wavelength

Explanation:

Electromagnetic Waves have different wavelengths.

They only differ from each other in wavelength. Wavelength is the distance between one wave crest to the next.

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A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen
MrMuchimi

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg 


The frequency of mass-spring oscillation is: 
f = (1/2π)√(k/m) 
k = m(2πf)² 

Then we know that k is constant for both trials, we have: 
k = k 


m1(2πf1)² = m2(2πf2)² 

m1 = m2(f2/f1)² 


m1 = (m1+Δm)(f2/f1)² 


m1 = Δm/((f1/f2)²-1)

 m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

5 0
3 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X

^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X

^{211}_{83}Bi\Rightarrow ^{211}_{84}Po

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

^{22}_{11}Na\Rightarrow ^{22}_{11-1}X

^{22}_{11}Na\Rightarrow ^{22}_{10}Ne

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

4 0
3 years ago
Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for
Zanzabum

Calm, sunny days with wind moving away from the center.

5 0
3 years ago
Read 2 more answers
As longitudional waves travel, particles in the medium are pushed together and then pulled apart. We call this
andreyandreev [35.5K]
Compression and rarefraction, the other guy's answer it's wrong
4 0
3 years ago
Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th
dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

         

Let's write the diffraction equation for a slit

       a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

             θ = λ / a

In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

           θ = 1.22 λ / D

Where D is the diameter of the pupil

 Let's apply this equation to our case

        θ = 1.22 600 10⁻⁹ / 0.460 10⁻²

        θ = 1.591 10⁻² rad

This is the angle separation to solve the two light sources

6 0
3 years ago
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